# How to Find Tangent Lines

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A tangent line to a curve touches the curve at only one point, and its slope is equal to the slope of the curve at that point. You can estimate the tangent line using a kind of guess-and-check method, but the most straightforward way to find it is through calculus. The derivative of a function gives you its slope at any point, so by taking the derivative of the function that describes your curve, you can find the slope of the tangent line then solve for the other constant to get your answer.

Write down the function for the curve whose tangent line you need to find. Determine at which point you want to take the tangent line (e.g., x = 1).

Take the derivative of the function using the derivative rules. There are too many to summarize here; you can find a list of the rules of derivation under the Resources section, however, in case you need a refresher:

Example: If the function is f(x) = 6x^3 + 10x^2 - 2x + 12, the derivative would be as follows:

f'(x) = 18x^2 + 20x - 2

Note that we represent the derivative of the original function by adding the ' mark, so that f'(x) is the derivative of f(x).

Plug the x-value for which you need the tangent line into f'(x) and calculate what f'(x) will be at that point.

Example: If f'(x) is 18x^2 + 20x - 2 and you need the derivative at the point where x = 0, then you would plug 0 into this equation in place of x to obtain the following:

f'(0) = 18 (0)^2 + 20(0) - 2

so f'(0) = -2.

Write out an equation of the form y = mx + b. This will be your tangent line. m is the slope of your tangent line and it's equal to your result from step 3. You don't know b yet, however, and will need to solve for it. Continuing the example, your initial equation based on step 3 would be y = -2x + b.

Plug the x-value you used to find the slope of the tangent line back into your original equation, f(x). This way, you can determine the y-value of your original equation at this point, then use it to solve for b in your tangent line equation.

Example: If x is 0, and f(x) = 6x^3 + 10x^2 - 2x + 12, then f(0) = 6(0)^3 + 10(0)^2 - 2(0) + 12. All terms in this equation go to 0 except for the last one, so f(0) = 12.

Substitute the result from step 5 for y in your tangent line equation, then substitute the x-value you used in step 5 for x in your tangent line equation and solve for b.

Example: You know from a prior step that y = -2x + b. If y = 12 when x = 0, then 12 = -2(0) + b. The only possible value for b that will give a valid result is 12, therefore b = 12.

Write out your tangent line equation, using the m and b values you have found.

Example: You know m = -2 and b = 12, so y = -2x + 12.

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